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Unformatted text preview: (1.7 m/s ) 2 Substituting this result into Equation (2) gives
2
m (1.7 m/s ) = 1 mv0
2
2 85. or v0 = 2 (1.7 m/s ) = 2.4 m/s
2 SSM WWW REASONING Gravity is the only force acting on the vaulters, since
friction and air resistance are being ignored. Therefore, the net work done by the
nonconservative forces is zero, and the principle of conservation of mechanical energy
holds.
SOLUTION Let E2f represent the total mechanical energy of the second vaulter at the
ground, and E20 represent the total mechanical energy of the second vaulter at the bar.
Then, the principle of mechanical energy is written for the second vaulter as
1
1
2
2
mv + mgh = mv20 + mgh
2 42f2442f 2 4424420
14
31
3
E2 f
E20 Since the mass m of the vaulter appears in every term of the equation, m can be eliminated
algebraically. The quantity h20 is h20 = h, where h is the height of the bar. Furthermore,
when the vaulter is at ground level, h2f = 0 m. Solving for v20 we have
2
v 20 = v 2 f − 2 gh (1) 338 WORK AND ENERGY In order to use Equation (1), we must first determine the height h of the bar. The height h
can be determined by applying the principle of conservation of mechanical energy to the
first vaulter on the ground and at the bar. Using notation similar to that above, we have
1
1
2
2
mv1f + mgh1f = mv10 + mgh10
24
2
14 244
3 144244
3
E1f
E10 Where E10 corresponds to the total mechanical energy of the first vaulter at the bar. The
height of the bar is, therefore, v2 − v2
(8.90 m/s)2 − (1.00 m/s)2
h = h10 = 1f 10 =
= 3.99 m
2g
2(9.80 m/s2 )
The speed at which the second vaulter clears the bar is, from Equation (1),
2 2 2 v20 = v2f − 2 gh = (9.00 m/s) − 2(9.80 m/s )(3.99 m) = 1.7 m/s
______________________________________________________________________________
86. REASONING AND SOLUTION At the bottom of the circular path of the swing, the
centripetal force is provided by the tension in the rope less the weight of the swing and
rider. That is,
2 mv
= T − mg
r
{
FC Solving for the mass yields
m= T
2 v
+g
r The energy of the swing is conserved if friction is ignored. The initial energy, E0, when the
swing is released is completely potential energy and is E0 = mgh0, where
h0 = r (1 – cos 60.0°) = 1r
2 is the vertical height of the swing. At the bottom of the path the swing's energy is entirely
kinetic and is
Ef = 1 mv
2 Setting Eo = Ef and solving for v gives 2 Chapter 6 Problems 339 v = 2gh0 = gr The expression for the mass now becomes 8.00 × 102 N
T
=
= 40.8 kg
2 g 2 9.80 m/s 2
______________________________________________________________________________
m= ( ) 87. REASONING The magnitude fk of the average kinetic friction force exerted on the
skateboarder is related to the amount of work Wk the done by that force and the distance s he slides along the ramp according to Wk = ( f k cos θ ) s (Equation 6.1). Solving this equation
for the kinetic friction force, we find fk = Wk (1) s cos θ The ramp is semicircular, so sliding from the top to the bottom means sliding onefourth of
the circumference of a circle with a radius of r = 2.70 m. Therefore, s = 1 ( 2π r ) = 1 π r .
4
2
Because the kinetic friction force is always directed opposite the skateboarder’s velocity, the
angle θ in Equation (1) is 180°. Thus, Equation (1) becomes fk =
To complete Wnc = the Wk
1 π r cos180
2 solution, o we =− 2Wk (2) πr will use the workenergy theorem ( 1 mvf2 − 1 mv02 ) + ( mghf − mgh0 ) (Equation 6.6) to find the work Wk done by the
2
2 nonconservative kinetic friction force. We can do this because the normal force exerted by
the ramp, which is also a nonconservative force, is always perpendicular to the
skateboarder’s velocity. Consequently, the normal force does no work, so the net work Wnc
done by all nonconservative forces is just the work Wk done by the kinetic friction force: Wnc = Wk = ( 12 mvf2 − 1 mv02 ) + ( mghf − mgh0 ) = 1 m ( vf2 − v02 ) + mg ( hf − h0 )
2
2 (3) SOLUTION Substituting Equation (3) into Equation (2), we obtain ( ) 2
2 1 m vf2 − v0 + mg ( hf − h0 ) 2 =−
fk = −
πr
πr 2Wk (4) 340 WORK AND ENERGY The skateboarder falls from rest, so we have v0 = 0 m/s. He starts a distance h above the top
of the ramp, which itself is a distance r above the bottom of the ramp. His vertical
displacement, therefore, is hf − h0 = − ( h + r ) . The algebraic sign of this displacement is
negative because he moves downward. Suppressing units, Equation (4) then yields ( ) 2
2
2 1 m vf2 − 0 − mg ( h + r ) 2 = − mvf − 2mg ( h + r ) = 2mg ( h + r ) − mvf
fk = − πr
πr
πr (5) Thus, the average force of kinetic friction acting on the skateboarder as he slides down the
ramp is
fk = ( ) 2 ( 61.0 kg ) 9.80 m/s 2 (1.80 m + 2.70 m ) − ( 61.0 kg )( 6.40 m/s ) π ( 2.70 m ) 2 = 3.40 × 10 2 N CHAPTER 7 IMPULSE AND MOMENTUM
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
_____________________________________________________________________________________________ 1. (b) Kinetic energy, 1 mv 2 , is a scalar quantity and is...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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